Note that proving \(\mathsf {CG}^{\text{pub}}(f) = O(\mathsf {CG}^{\text{pub}}_{[1]}(f)^2)\)  would prove that \(\mathsf {bs}(f) \le O(\mathsf {s}(f)^2)\) .
It may seem too much to expect that the single-bit version
of the game can help get upper bounds on the general public coin setting, but thanks to Huang's breakthrough result [1]}, we already know that
\(\mathsf {CG}^{\text{pub}}(f) = O(\mathsf {CG}^{\text{pub}}_{[1]}(f)^5)\)  for any total Boolean function \(f\) .
